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# A hydrogen atom, unipositive helium, and dipositive lithium contain a single electron in the same shell. The radius of the shell:A.Decreases B.Increases C.Remains unaffectedD.Cannot be predicted

Last updated date: 15th Aug 2024
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Hint: For this question we must have the knowledge of the Bohr model for hydrogen and hydrogen-like species. We can calculate the radius of the shell by using the atomic number of their respective species in the formula.

Formula used: ${{\text{R}}_{\text{n}}} = {{\text{R}}_ \circ } \times \dfrac{{{{\text{n}}^2}}}{{\text{Z}}}$ where ${{\text{R}}_{\text{n}}}$ is radius of ${{\text{R}}_{\text{n}}}$ orbital ${{\text{R}}_{\text{n}}}$ is Bohr radius that is constant, n is number of orbit and s always a natural number and Z is atomic number of element.

Complete step by step solution:
We need to check the variation of radius for the given species. Let us calculate the radius of atom in each case:
The atomic number of hydrogen is 1. Hence ${\text{Z }} = 1$ and we need to calculate for ${\text{n }} = 1$. Putting the values in the formula we will get:
${{\text{R}}_{1({\text{H)}}}} = {{\text{R}}_ \circ } \times \dfrac{{{1^2}}}{1} = {{\text{R}}_ \circ }$
The atomic number of Helium is 2. Helium ion is ${\text{H}}{{\text{e}}^ + }$, but the atomic number remains the same whether it is an ion or atom of the same elements. Hence ${\text{Z }} = 2$ and we need to calculate for ${\text{n }} = 1$. Putting the values in the formula we will get:
${{\text{R}}_{1({\text{H}}{{\text{e}}^ + }{\text{)}}}} = {{\text{R}}_ \circ } \times \dfrac{{{1^2}}}{2} = \dfrac{{{{\text{R}}_ \circ }}}{2}$
The atomic number of Lithium is 3. Dipositive Lithium ion is ${\text{L}}{{\text{i}}^{2 + }}$, but the atomic number remains same whether it is an ion or atom of same elements. Hence ${\text{Z }} = 3$ and we need to calculate for ${\text{n }} = 1$. Putting the values in the formula we will get:
${{\text{R}}_{1({\text{L}}{{\text{i}}^{ + 2}}{\text{)}}}} = {{\text{R}}_ \circ } \times \dfrac{{{1^2}}}{3} = \dfrac{{{{\text{R}}_ \circ }}}{3}$
Hence the radius of ${\text{H, H}}{{\text{e}}^ + }{\text{and L}}{{\text{i}}^{ + 2}}$ is ${{\text{R}}_{\text{o}}}{\text{, }}\dfrac{{{{\text{R}}_{\text{o}}}}}{2},{\text{ }}\dfrac{{{{\text{R}}_{\text{o}}}}}{3}$ respectively. Hence, we can clearly see that radius decreases.
The correct option is A.

Note: The given species ${\text{H, H}}{{\text{e}}^ + }{\text{and L}}{{\text{i}}^{ + 2}}$ are iso-electronic in nature. That is all these species have same number of electron though have different atomic number. Hydrogen has 1 electron and 1 proton. Helium has atomic number 2 and have 2 electrons but in ${\text{H}}{{\text{e}}^ + }$ has 1 less electron due to positive charge that makes 1 electron in ${\text{H}}{{\text{e}}^ + }$ and similarly 1 electron is there in ${\text{L}}{{\text{i}}^{2 + }}$ sue to 2 positive charge.